Losing my marbles.... Probability

OK so I have a bag of marbles containing: 3 Catseyes, 4 Agates and 7 Oxbloods. If all have an equal chance of being picked and two are drawn, what are the chances of at least 1 of the chosen being an Agate.
According to my limited learning I worked it out as follows:

$\displaystyle P(at least 1 agate)= \frac{4}{14} + \frac{4}{13} = \frac{54}{91} $
Is this correct? If not then how do I work it out?

OK so I have a bag of marbles containing: 3 Catseyes, 4 Agates and 7 Oxbloods. If all have an equal chance of being picked and two are drawn, what are the chances of at least 1 of the chosen being an Agate.
According to my limited learning I worked it out as follows:

$\displaystyle P(at least 1 agate)= \frac{4}{14} + \frac{4}{13} = \frac{54}{91} $
Is this correct? If not then how do I work it out?